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Python最长公共子串算法实例

2015-03-07 10:09 736 查看

本文实例讲述了Python最长公共子串算法。分享给大家供大家参考。具体如下：

#!/usr/bin/env python
# find an LCS (Longest Common Subsequence).
# *public domain*
def find_lcs_len(s1, s2):
m = [ [ 0 for x in s2 ] for y in s1 ]
for p1 in range(len(s1)):
for p2 in range(len(s2)):
if s1[p1] == s2[p2]:
if p1 == 0 or p2 == 0:
m[p1][p2] = 1
else:
m[p1][p2] = m[p1-1][p2-1]+1
elif m[p1-1][p2] < m[p1][p2-1]:
m[p1][p2] = m[p1][p2-1]
else:               # m[p1][p2-1] < m[p1-1][p2]
m[p1][p2] = m[p1-1][p2]
return m[-1][-1]
def find_lcs(s1, s2):
# length table: every element is set to zero.
m = [ [ 0 for x in s2 ] for y in s1 ]
# direction table: 1st bit for p1, 2nd bit for p2.
d = [ [ None for x in s2 ] for y in s1 ]
# we don't have to care about the boundery check.
# a negative index always gives an intact zero.
for p1 in range(len(s1)):
for p2 in range(len(s2)):
if s1[p1] == s2[p2]:
if p1 == 0 or p2 == 0:
m[p1][p2] = 1
else:
m[p1][p2] = m[p1-1][p2-1]+1
d[p1][p2] = 3          # 11: decr. p1 and p2
elif m[p1-1][p2] < m[p1][p2-1]:
m[p1][p2] = m[p1][p2-1]
d[p1][p2] = 2          # 10: decr. p2 only
else:               # m[p1][p2-1] < m[p1-1][p2]
m[p1][p2] = m[p1-1][p2]
d[p1][p2] = 1          # 01: decr. p1 only
(p1, p2) = (len(s1)-1, len(s2)-1)
# now we traverse the table in reverse order.
s = []
while 1:
print p1,p2
c = d[p1][p2]
if c == 3: s.append(s1[p1])
if not ((p1 or p2) and m[p1][p2]): break
if c & 2: p2 -= 1
if c & 1: p1 -= 1
s.reverse()
return ''.join(s)
if __name__ == '__main__':
print find_lcs('abcoisjf','axbaoeijf')
print find_lcs_len('abcoisjf','axbaoeijf')

希望本文所述对大家的Python程序设计有所帮助。

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